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Maslowich
3 years ago
7

A gas that exerts a pressure of 15.6 psi in a container with the volume of ______ L will exert a pressure of 24.43 psi when tran

sferred to a container with a volume of 1.895 L.
Chemistry
2 answers:
dybincka [34]3 years ago
4 0
Boyle's Law states: pV = constant.

24.43 x 1.895 = 46.29485

therefore, 15.6 x _____ = 46.29485

unknown = 2.968L
Katyanochek1 [597]3 years ago
4 0

Answer:

The initial volume of the container = 2.97 L

Explanation:

<u>Given:</u>

Initial pressure of gas, P1 = 15.6 psi

Final pressure of gas, P2 = 24.43 psi

Final volume, V2 = 1.895 L

<u>To determine:</u>

The initial volume V1 occupied by the gas

<u>Explanation:</u>

Based on the ideal gas equation

PV = nRT

where P = pressure, V = volume ; n = moles of gas

R = gas constant, T = temperature

At constant n and T, the above equation becomes:

PV = constant

This is the Boyle's law

Therefore:

P1V1 = P2V2\\\\V1 = \frac{P2V2}{P1} = \frac{24.43 \ psi*1.895\ L}{15.6\ psi} =2.97\ L

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wolverine [178]

Answer:

6 atoms

Explanation:

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3 years ago
If you can plz help ASAP Read the chemical equation.
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in this question, we are dealing with only NH3 and H2 so we only focus on that

since the ratio of H2 to 2NH3 is 3:2, we say that

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3.6 litres of H2 = x liters of 2NH3

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6 0
2 years ago
Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the
GarryVolchara [31]

Answer : The final temperature of the metal block is, 25^oC

Explanation :

heat_{absorbed}=heat_{released}

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Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of aluminum = 55 g

m_2 = mass of water = 0.48 g

T_{final} = final temperature = ?

T_1 = temperature of aluminum = 25^oC

T_2 = temperature of water = 25^oC

c_1 = specific heat of aluminum = 0.900J/g^oC

c_2 = specific heat of water= 4.184J/g^oC

Now put all the given values in equation (1), we get

55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]

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Energy of motion is referred to as CA. thermal energy B. electrical energy C. potential energy D. kinetic energy​
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8 0
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Answer:

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7 0
2 years ago
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